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Neko [114]
3 years ago
10

a crane lifts a 75 kg mass at a height of 8m . calculate the gravitational potential energy gained by the mass (g= 10N)

Physics
1 answer:
Bogdan [553]3 years ago
5 0
Advice: Search up your questions in the Brainly search bar before asking your questions.
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3 years ago
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A 14.0 gauge copper wire of diameter 1.628 mmcarries a current of 14.0 mA. Part A: What is the potential difference across a 2.1
FrozenT [24]

Answer:

for copper potential difference is 2.4871 × 10^{-4} V

for silver  potential difference is 2.1256 × 10^{-4} V

Explanation:

given data

diameter D = 1.628 mm = 0.001628 m

current = 14.0 mA = 0.0140 A

length L = 2.15 m

to find out

potential difference

solution

we consider here

resistivity of silver =  1.47 × 10^{-8} ohm-m

resistivity of copper = 1.72 × 10^{-8} ohm-m

so we apply here resistance formula that is

Resistance = ρ × L / A     ...............1

so here area A = πr² = π(0.001628 /2)² = 2.0816 ×  10^{-6} m²

for copper Resistance = ρ × L / A

Resistance = 1.72 × 10^{-8}  × 2.15 / 2.0816 ×  10^{-6}

Resistance = 1.7765 × 10^{-2}  ohm

for silver Resistance = ρ × L / A

Resistance = 1.47 × 10^{-8}  × 2.15 / 2.0816 ×  10^{-6}

Resistance = 1.5183 × 10^{-2}  ohm

so

potential difference  is calculated as

for copper potential difference  = current × resistance

potential difference  = 0.0140 ×1.7765 × 10^{-2}

potential difference  = 2.4871 × 10^{-4} V

for silver  potential difference  = current × resistance

potential difference  =  0.0140 × 1.5183 × 10^{-2}

potential difference  =  2.1256 × 10^{-4} V

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What charge are electrons
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The force exerted by an electric charge at the origin on a charged particle at a point (x, y, z) with position vector r = x, y,
kap26 [50]

W = K[ {\frac{1}{3} - \frac{1}{\sqrt{38} }  ]

<u>Explanation:</u>

The parametric representation of a line segment joining the points (a,b,c) and (l,m,n) is

r(t) = (1-t) . (a,b,c) + t . (l, m, n)  where t ∈ |0, 1|

So, the parametric representation of a line segment joining the points (3,0,0) and (3,2,5) is

r(t) = (1 - t) . (3,0,0) + t . (3,2,5)  where t ∈ |0, 1|

r(t) = (3(1 - t), 0, 0) + (3t, 2t, 5t)  where t ∈ |0, 1|

r(t) = (3, 2t, 5t)

Given:

F(x, y, z) = \frac{Kr}{|r|^3} \\\\F(x, y, z) = \frac{K}{(x^2 + y^2 + z^2)^3^/^2}  (x, y, z)\\\\F(r(t)) = \frac{K}{(3^2 + (2t)^2 + (5t)^2)^3^/^2}  (3, 2t, 5t)\\\\F(r(t)) = \frac{K}{(9 + 29t^2)^3^/^2} (3, 2t, 5t)

dr = (0, 2, 5) dt

Work = \int\limits^1_0 {F} \, dr

W = \int\limits^1_0 {\frac{K}{(9 + 29t^2)^3^/^2} } (3, 2t, 5t) . (0, 2, 5)\, dt\\ \\    = \int\limits^1_0 {\frac{K ( 4t + 25t)}{(9 + 29t^2)^3^/^2} } \, dt\\\\\\

W = \frac{1}{2}\int\limits^1_0 {\frac{K(29t)}{(9 + 29t^2)^3^/^2} } \, dt \\ \\

Substitute = 9 + 29t² = u, 92tdt = du

Limit changes from 0→1 to 9 → 38

W = \frac{K}{2} \int\limits^3_9 {\frac{du}{u^3^/^2} } \,\\\\

On solving this, we get:

W = K[ {\frac{1}{3} - \frac{1}{\sqrt{38} }  ]

Therefore, work done is W = K[ {\frac{1}{3} - \frac{1}{\sqrt{38} }  ]

5 0
3 years ago
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